# How can you add pressure and internal energy in relativistic enthalpy?

In relativistic fluid dynamics the relativistic enthalpy in natural units is defined as: $$\begin{equation} h = \frac{e+p}{\rho}, \end{equation}$$ Where $$e$$ is the total energy density and $$p$$ is the thermodynamic pressure. In contrast, the Newtonian enthalpy is defined as: $$\begin{equation} h_N = \epsilon+\frac{p}{\rho}, \end{equation}$$ Where $$\epsilon$$ is the specific internal energy and $$\rho$$ is the rest mass density.

The question: What are the units of $$e$$ and $$p$$ so that I can add them in the definition of $$h$$ above?

• As you said, they are in natural units where $c=1$. Jan 14 at 18:49
• True, but it's still very confusing. For example, is it the energy that has units of pressure, or is it the pressure that has units of energy, or am I just missing $c$ or $\hbar$ factor somewhere? How come energy and pressure have the same units in the first place? They're completely different physical quantities (ok, this would be explained by the $c=\hbar=1$ factor). Jan 14 at 19:23
• There are missing powers of $c$ (as is common when doing Relativity). I don’t remember where they go, but it would be good practice for you to use dimensional analysis to figure this out. Jan 14 at 19:43
• I’ve led you astray. Energy density and pressure have the same dimensions $ML^{-1}T^{-2}$ without taking $c=1$. Jan 14 at 19:59
• The dimensions of torque and energy also match, even though they are conceptually different. Jan 14 at 23:14

So, following @Ghoster's comments, $$e$$ and $$p$$ simply have the same dimensions in both systems of units: $$\begin{equation} e = \underbrace{(M\ L^2\ T^{-2})}_{energy} \cdot \underbrace{L^{-3}}_{per\ unit\ volume} = \left( \underbrace{(M \ L \ T^{-2})}_{force} \cdot \underbrace{L}_{times\ length} \right) \cdot \underbrace{L^{-3}}_{per\ unit\ volume} = M\ L^{-1}\ T^{-2} \\ p = \underbrace{(M \ L \ T^{-2})}_{force} \cdot \underbrace{L^{-2}}_{per\ unit\ area} = M\ L^{-1}\ T^{-2} = e \end{equation}$$